java.lang.RuntimeException:Handler(android.os.Handler)在死线程上向Handler发送消息

Question

在我的应用程序中我使用IntentService发送短信.

    @Override
protected void onHandleIntent(Intent intent) {
    Bundle data = intent.getExtras();
    String[] recipients = null;
    String message = getString(R.string.unknown_event);
    String name = getString(R.string.app_name);
    if (data != null && data.containsKey(Constants.Services.RECIPIENTS)) {
        recipients = data.getStringArray(Constants.Services.RECIPIENTS);
        name = data.getString(Constants.Services.NAME);
        message = data.getString(Constants.Services.MESSAGE);
        for (int i = 0; i < recipients.length; i++) {
            if(!StringUtils.isNullOrEmpty(recipients[i])) {
                try {
                    Intent sendIntent = new Intent(this, SMSReceiver.class);
                    sendIntent.setAction(Constants.SMS.SEND_ACTION);
                    PendingIntent sendPendingIntent = PendingIntent.getBroadcast(getApplicationContext(), 0, sendIntent, PendingIntent.FLAG_UPDATE_CURRENT);
                    Intent deliveryIntent = new Intent(this, SMSReceiver.class);
                    deliveryIntent.setAction(Constants.SMS.DELIVERED_ACTION);
                    PendingIntent deliveryPendingIntent = PendingIntent.getBroadcast(getApplicationContext(), 0, deliveryIntent, PendingIntent.FLAG_UPDATE_CURRENT);
                    SmsManager.getDefault().sendTextMessage(recipients[i].trim(), null, "[" + name + "] " + message, sendPendingIntent, deliveryPendingIntent);
                } catch (Exception e) {
                    Log.e(TAG, "sendTextMessage", e);
                    e.printStackTrace();
                    Toast.makeText(this, e.getMessage(), Toast.LENGTH_LONG).show();
                    MainActivity.instance.writeToLogFile(e.getMessage(), System.currentTimeMillis());                       
                }
            }
        }
    }
}

运行应用程序时,我收到以下错误:

W/MessageQueue(7180): Handler (android.os.Handler) {42586468} sending message to a Handler on a dead thread
W/MessageQueue(7180): java.lang.RuntimeException: Handler (android.os.Handler) {42586468} sending message to a Handler on a dead thread
W/MessageQueue(7180):   at android.os.MessageQueue.enqueueMessage(MessageQueue.java:294)
W/MessageQueue(7180):   at android.os.Handler.enqueueMessage(Handler.java:618)
W/MessageQueue(7180):   at android.os.Handler.sendMessageAtTime(Handler.java:587)
W/MessageQueue(7180):   at android.os.Handler.sendMessageDelayed(Handler.java:558)
W/MessageQueue(7180):   at android.os.Handler.post(Handler.java:323)
W/MessageQueue(7180):   at android.widget.Toast$TN.hide(Toast.java:367)
W/MessageQueue(7180):   at android.app.ITransientNotification$Stub.onTransact(ITransientNotification.java:55)
W/MessageQueue(7180):   at android.os.Binder.execTransact(Binder.java:351)
W/MessageQueue(7180):   at dalvik.system.NativeStart.run(Native Method)

我的SMSReceiver位于另一个类中.我该如何解决这个问题？谢谢; 的Eyal.

Answer 1

这里的问题是你正在创建一个Toast由a管理的线程内部IntentService.系统将使用Handler与此线程关联来显示和隐藏Toast.

首先Toast会正确显示,但是当系统试图隐藏它时,在onHandleIntent方法完成后,将抛出"在死线程上向处理程序发送消息"的错误,因为Toast创建它的线程不再有效,并Toast不会消失.

为避免这种情况,您应该显示Toast向主线程发布消息.这是一个例子:

    // create a handler to post messages to the main thread
    Handler mHandler = new Handler(getMainLooper());
    mHandler.post(new Runnable() {
        @Override
        public void run() {
            Toast.makeText(getApplicationContext(), "test", Toast.LENGTH_SHORT).show();
        }
    });

Answer 2

在IntentService上显示Toast。试试这个代码..

@Override
public void onCreate() {
    super.onCreate();
    mHandler = new Handler();
}

@Override
protected void onHandleIntent(Intent intent) {
    mHandler.post(new Runnable() {            
        @Override
        public void run() {
            Toast.makeText(MyIntentService.this, "Test", Toast.LENGTH_LONG).show();                
        }
    });
}

来源：-https : //stackoverflow.com/a/5420929/4565853